(January 2010 #4) Suppose that \(f:\mathbb R\to\mathbb R\) is differentiable at \(a\in\mathbb R\). If \(\{x_n\}\) is an increasing sequence of real numbers converging to \(a\) and \(\{y_n\}\) is a decreasing sequence of real numbers converging to \(a\), prove that \[\lim_{n\to\infty}\frac{f(y_n)-f(x_n)}{y_n-x_n}=f'(a).\]
Solution
1. Decompose the quotient.Proof: write \(x_n \nearrow a\), \(y_n \searrow a\) with \(x_n < a < y_n\) for all \(n\).Then \[\frac{f(y_n) - f(x_n)}{y_n - x_n} = \frac{(f(y_n) - f(a)) + (f(a) - f(x_n))}{y_n - x_n} = \frac{y_n - a}{y_n - x_n}\cdot\frac{f(y_n) - f(a)}{y_n - a} + \frac{a - x_n}{y_n - x_n}\cdot\frac{f(a) - f(x_n)}{a - x_n}.\]
2. The coefficients are a convex combination.Proof: \(\lambda_n = \frac{y_n - a}{y_n - x_n} \in (0,1)\) and \(\frac{a - x_n}{y_n - x_n} = 1 - \lambda_n\); indeed \((y_n - a) + (a - x_n) = y_n - x_n\).
3. Both difference quotients tend to \(f'(a)\).Proof: \(\frac{f(y_n) - f(a)}{y_n - a} \to f'(a)\) as \(y_n \to a^+\) (subsequence of the defining limit), and \(\frac{f(a) - f(x_n)}{a - x_n} = \frac{f(x_n) - f(a)}{x_n - a} \to f'(a)\) as \(x_n \to a^-\).
4. Conclude.Proof: the quotient is \(\lambda_n A_n + (1-\lambda_n)B_n\) with \(A_n, B_n \to f'(a)\) and \(0 \le \lambda_n \le 1\); hence \[\lim_{n\to\infty}\frac{f(y_n) - f(x_n)}{y_n - x_n} = f'(a).\]
5. Q.E.D.